\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 27, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/27\hfil Regularized trace]
{Regularized trace of the Sturm-Liouville operator
with irregular boundary conditions}

\author[A. Makin\hfil EJDE-2009/27\hfilneg]
{Alexander Makin}

\address{Alexander Makin \newline
Moscow State University of Instrument-Making and Informatics \newline
Stromynka 20, Moscow, 107996, Russia}
\email{alexmakin@yandex.ru}

\thanks{Submitted January 8, 2009. Published February 3, 2009.}
\subjclass[2000]{34L05, 34B24}
\keywords{Sturm-Liouville operator; eigenvalue problem; spectrum}

\begin{abstract}
 We consider the spectral problem for the
 Sturm-Liouville equation with a complex-valued potential $q(x)$
 and with irregular boundary conditions on the interval $(0,\pi)$.
 We establish a formula for the first regularized trace of the
 this operator.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks

\section{Introduction and main result}

This paper deals with the eigenvalue problem for the
Sturm-Liouville equation
\begin{equation}
u''-q(x)u+\lambda u=0 \label{e1.1}
\end{equation}
on the interval $(0,\pi)$
with the boundary conditions
\begin{equation}
u'(0)+(-1)^\theta u'(\pi)+bu(\pi)=0,\quad
u(0)+(-1)^{\theta+1}u(\pi)=0,\label{e1.2}
\end{equation}
where $b$ is a complex number, $b\ne0$, $\theta=0,1$.
The goal of this article is to calculate the first-order regularized
trace for  \eqref{e1.1}-\eqref{e1.2}.

The theory of regularized traces of ordinary differential
operators has a long history. First, the trace formulas for the
Sturm-Liouville operator with the Dirichlet boundary conditions
and sufficiently smooth potential $q(x)$ were established in
\cite{d1,g1}.
 Afterwards these investigations were continued in many
directions, for instance, the trace formulas for the
Sturm-Liouville operator with periodic or antiperiodic boundary
conditions were obtained in \cite{l1,s2}, and for regular but not
strongly regular ones \cite{n1} similar formulas were found in
\cite{m2}. A
method for calculating trace formulas for general problems
involving ordinary differential equations on a finite interval was
proposed in \cite{l3}. The bibliography on the subject is very extensive
and we refer to the list of the works in \cite{l2,s1}.

The trace formulas can be used for approximate calculation of the
first eigenvalues of an operator \cite{s1}, and in order to establish
necessary and sufficient conditions for a set of complex numbers
to be the spectrum of an operator \cite{s2}.


We will prove the following statement.


\begin{theorem} \label{thm1}
Let $q(x)$ be an arbitrary complex-valued
function in $W_1^1(0,\pi)$ and denote by $\lambda_n$
$(n=1,2,\dots)$ the eigenvalues of  \eqref{e1.1}-\eqref{e1.2}. Then we
have the trace formula
\begin{equation}
\begin{aligned}
&\sum_{n=1}^\infty\big\{\lambda_n-n^2-\langle q\rangle
-\frac{(-1)^{\theta}(q(\pi)-q(0)-\int_0^\pi q'(t)\cos(2nt)dt)}{\pi b}\big\}\\
&-\frac{\langle q\rangle}{2}+\frac{(q(\pi)-q(0))^2}{2b^2}+\frac{q(\pi)+q(0)}{4}=0,
\end{aligned}\label{e1.3}
\end{equation}
where $\langle q\rangle=\pi^{-1}\int_0^\pi q(t)dt$.
\end{theorem}


This article is organized as follows. Section 2 is devoted to the analysis
of the characteristic determinant. In section 3, we obtain a formula of the
regularized trace in explicit form.

\section{Analysis of the characteristic determinant}

Denote by $c(x,\mu), s(x,\mu)$ $(\mu^2=\lambda)$ the fundamental
system of solutions to  \eqref{e1.1} with the initial conditions
$c(0,\mu)=s'(0,\mu)=1$, $c'(0,\mu)= s(0,\mu)=0$. Simple
calculations show that the characteristic equation of
\eqref{e1.1}-\eqref{e1.2} can be reduced to the form $\Delta(\mu)=0$, where
\begin{equation}
\Delta(\mu)=c(\pi,\mu)-s'(\pi,\mu)-(-1)^\theta bs(\pi,\mu).\label{e2.1}
\end{equation}
Denote by $\varphi(x,\mu), \psi(x,\mu)$
the system of solutions to  \eqref{e1.1}
with the initial conditions $\varphi(0,\mu)=\psi(0,\mu)=1$,
 $\varphi_x'(0,\mu)=i\mu$,
$\psi_x'(0,\mu)=-i\mu$. It is readily seen that
\begin{equation}
\begin{gathered}
\varphi(x,\mu)=\psi(x,-\mu),\quad
 c(x,\mu)=(\varphi(x,\mu)+\psi(x,\mu))/2,\\ s(x,\mu)=(\varphi(x,\mu)-\psi(x,\mu))/(2i\mu),\\
s'(x,\mu)=(\varphi'(x,\mu)-\psi'(x,\mu))/(2i\mu).
\end{gathered}\label{e2.2}
\end{equation}
For convenience, we introduce $I(x)=\int_0^x q(t)dt$. Asymptotic
formulas for the functions $\varphi(x,\mu)$ and $\varphi'(x,\mu)$
were established in \cite{m2}:
\begin{equation}
\begin{aligned}
\varphi(x,\mu)&=e^{i\mu x}[1+\frac{1}{2i\mu}I(x)-\frac{1}{8\mu^2}I^2(x)]\\
&\quad -\frac{e^{-i\mu x}}{2i\mu}\int_0^xe^{2i\mu t}q(t)dt+A(x,\mu)+R_1(x,\mu),
\end{aligned}\label{e2.3}
\end{equation}
and
\begin{equation}
\begin{aligned}
\varphi_x'(x,\mu)&=i\mu e^{i\mu x}[1+\frac{1}{2i\mu}I(x)
-(8\mu^2)^{-1}I^2(x)]\\
&\quad +\frac{e^{-i\mu x}}{2}\int_0^xe^{2i\mu t}q(t)dt+B(x,\mu)
+R_3(x,\mu)+R_4(x,\mu)+R_5(x,\mu),
\end{aligned}\label{e2.4}
\end{equation}
where
\begin{align*}
A(x,\mu)&= (4\mu^2)^{-1}(e^{-i\mu x}\int_0^xe^{2i\mu t}q(t)dt\int_0^tq(s)ds\\
&\quad +e^{i\mu x}\int_0^xe^{-2i\mu t}q(t)dt\int_0^te^{2i\mu s}q(s)ds
-e^{-i\mu x}\int_0^xq(t)dt\int_0^te^{2i\mu s}q(s)ds),
\end{align*}
\begin{equation}
\begin{aligned}
R_1(x,\mu)&= -(4\mu^3)^{-1}\int_0^x(e^{i\mu(x-t)}-e^{-i\mu(x-t)})q(t)dt\\
&\quad\times \int_0^t(e^{i\mu(t-s)}-e^{-i\mu(t-s)})q(s)ds
\int_0^s\sin\mu(s-y)q(y)\varphi(y,\mu)dy,
\end{aligned}
\label{e2.5}
\end{equation}
\begin{align*}
B(x,\mu)&= -(4i\mu)^{-1}[e^{i\mu x}\int_0^xe^{-2i\mu t}q(t)dt
 \int_0^te^{2i\mu s}q(s)ds\\
&\quad -e^{-i\mu x}\int_0^xe^{2i\mu t}q(t)dt\int_0^tq(s)ds
+e^{-i\mu x}\int_0^xq(t)dt\int_0^te^{2i\mu s}q(s)ds],
\end{align*}
\begin{gather}
R_2(x,\mu)=
-\frac{1}{16\mu^2}e^{i\mu x}\int_0^x(1+e^{2i\mu(t-x)})q(t)I^2(t)dt,\label{e2.6}
\\
R_3(x,\mu)=
\frac{1}{2}\int_0^x(e^{i\mu(x-t)}+e^{-i\mu(x-t)})q(t)R_2(t,\mu)dt,\label{e2.7}
\\
R_4(x,\mu)=
\frac{1}{2}\int_0^x(e^{i\mu(x-t)}+e^{-i\mu(x-t)})q(t)A(t,\mu)dt.\label{e2.8}
\end{gather}

We need more precise asymptotic formulas for the functions $\varphi(x,\mu)$
and $\varphi'(x,\mu)$.
Substituting known the expression \cite{m3} for the function
$\varphi(y,\mu)= e^{i\mu y}+O(1/\mu) e^{|Im\mu|y}$ into \eqref{e2.5}, we obtain
\begin{align*}
R_1(x,\mu)&=
-(8i\mu^3)^{-1}\int_0^x(e^{i\mu(x-t)}-e^{-i\mu(x-t)})q(t)dt\\
&\quad \times
\int_0^t(e^{i\mu(t-s)}-e^{-i\mu(t-s)})q(s)ds\int_0^s(e^{i\mu(s-y)}-e^{-i\mu(s-y)})q(y)\\
&\quad \times [e^{i\mu y}+O(1/\mu) e^{|Im\mu|y}]dy\\
&=-\frac{e^{i\mu x}}{8i\mu^3}\int_0^x(1-e^{-2i\mu(x-t)})q(t)dt
  \int_0^t(1-e^{-2i\mu(t-s)})q(s)ds\\
&\quad \times \int_0^s(1-e^{-2i\mu(s-y)})q(y)dy+ O(1/\mu^4) e^{|Im\mu|x}\\
&=-\frac{e^{i\mu x}}{8i\mu^3}\int_0^xq(t)dt\int_0^tq(s)ds\int_0^s q(y)\\
&\quad \times
[1-e^{-2i\mu(x-t)}-e^{-2i\mu(t-s)}-e^{-2i\mu(s-y)}+e^{-2i\mu(x-s)}\\
&\quad +e^{-2i\mu(x-t+s-y)}+e^{-2i\mu(t-y)}-e^{-2i\mu(x-y)}]dy
+O(1/\mu^4) e^{|Im\mu|x}\\
&=- \frac{e^{i\mu x}}{8i\mu^3}
\int_0^xq(t)dt\int_0^tq(s)ds\int_0^s q(y)dy
-\frac{1}{8i\mu^3}K(x,\mu)
+O(1/\mu^4) e^{|Im\mu|x},
\end{align*}
where
\begin{align*}
K(x,\mu)&=
e^{i\mu x}\int_0^xq(t)dt\int_0^tq(s)ds\int_0^s q(y)\\
&\quad \times
[-e^{-2i\mu(x-t)}-e^{-2i\mu(t-s)}-e^{-2i\mu(s-y)}+e^{-2i\mu(x-s)}\\
&\quad +e^{-2i\mu(x-t+s-y)}+e^{-2i\mu(t-y)}-e^{-2i\mu(x-y)}]dy.
\end{align*}
Arguing as in \cite{m1}, we see that
$$
\int_0^xq(t)dt\int_0^tq(s)ds\int_0^s q(y)dy
=\frac{1}{6}I^3(x).
$$
The obvious inequality $0\le y\le s\le t\le x$, together with the
inequality $|x-2z|\le x$
valid for $0\le z\le x$ and the Riemann lemma \cite{m3},
implies
\begin{equation}
K(x,\mu)=o(1)e^{|Im\mu|x},\quad
R_1(x,\mu)=-\frac{1}{48i\mu^3}I^3(x)+o(1/\mu^3) e^{|Im\mu|x}.\label{e2.9}
\end{equation}
Continuing this line of reasoning, we reduce \eqref{e2.6} and \eqref{e2.8} to
the  form
\begin{equation}
R_2(x,\mu)=-\frac{1}{48\mu^2}e^{i\mu x}I^3(x)+o(1/\mu^2) e^{|Im\mu|x},\quad
R_4(x,\mu)=o(1/\mu^2) e^{|Im\mu|x}.\label{e2.10}
\end{equation}
It follows from \eqref{e2.7} and \eqref{e2.9} that
\begin{equation}
R_3(x,\mu)=O(1/\mu^3) e^{|Im\mu|x}.\label{e2.11}
\end{equation}
Combining \eqref{e2.3} and \eqref{e2.9}, we obtain
\begin{equation}
\begin{aligned}
\varphi(x,\mu)&=e^{i\mu x}[1+\frac{1}{2i\mu}I(x)-\frac{1}{8\mu^2}I^2(x)]\\
&\quad -\frac{e^{-i\mu x}}{2i\mu}\int_0^xe^{2i\mu t}q(t)dt+A(x,\mu)-
\frac{e^{i\mu x}}{48i\mu^3}I^3(x)+o(1/\mu^3) e^{|Im\mu|x},
\end{aligned}\label{e2.12}
\end{equation}
It follows from \eqref{e2.4}, \eqref{e2.10} and \eqref{e2.11} that
\begin{equation}
\begin{aligned}
\varphi_x'(x,\mu)
&=i\mu e^{i\mu x}[1+\frac{1}{2i\mu}I(x)-(8\mu^2)^{-1}I^2(x)]\\
&\quad +\frac{e^{-i\mu x}}{2}\int_0^xe^{2i\mu t}q(t)dt+B(x,\mu)
-\frac{1}{48\mu^2}e^{i\mu x}I^3(x)+o(1/\mu^2) e^{|Im\mu|x}.
\end{aligned}\label{e2.13}
\end{equation}
Substituting into \eqref{e2.1} the expressions for $c(\pi,\mu)$,
$s(\pi,\mu)$, $s'(\pi,\mu)$ of \eqref{e2.2}, \eqref{e2.12}, \eqref{e2.13},
respectively, we obtain
\begin{align*}
&\Delta(\mu)\\
&=\frac{1}{2}\Big\{
e^{i\mu\pi}[1+\frac{1}{2i\mu}I(\pi)-\frac{1}{8\mu^2}I^2(\pi)]\\
&\quad -\frac{e^{-i\mu\pi}}{2i\mu}\int_0^\pi e^{2i\mu t}q(t)dt+A(\pi,\mu)-
\frac{e^{i\pi\mu}}{48i\mu^3}I^3(\pi)+o(1/\mu^3) e^{|Im\mu|\pi}\\
&\quad +e^{-i\mu\pi}[1-\frac{1}{2i\mu}I(\pi)-\frac{1}{8\mu^2}I^2(\pi)]\\
&\quad +\frac{e^{i\mu\pi}}{2i\mu}\int_0^\pi e^{-2i\mu t}q(t)dt+A(\pi,-\mu)+
\frac{e^{-i\pi\mu}}{48i\mu^3}I^3(\pi)+o(1/\mu^3) e^{|Im\mu|\pi}\Big\}\\
&\quad -\frac{1}{2i\mu}\Big\{
i\mu e^{i\mu\pi}[1+\frac{1}{2i\mu}I(\pi)
-(8\mu^2)^{-1}I^2(\pi)]\\
&\quad +\frac{e^{-i\mu\pi}}{2}\int_0^\pi e^{2i\mu t}q(t)dt+B(\pi,\mu)
-\frac{1}{48\mu^2}e^{i\mu\pi}I^3(\pi)+o(1/\mu^2) e^{|Im\mu|\pi}\\
&\quad -[-i\mu e^{-i\mu\pi}[1-\frac{1}{2i\mu}I(\pi)
-(8\mu^2)^{-1}I^2(\pi)]\\
&\quad +\frac{e^{i\mu\pi}}{2}\int_0^\pi e^{-2i\mu t}q(t)dt+B(\pi,-\mu)
-\frac{1}{48\mu^2}e^{-i\mu\pi}I^3(\pi)
+o(1/\mu^2) e^{|Im\mu|\pi}]\Big\}\\
&\quad -(-1)^\theta\frac{b}{2i\mu}\Big\{
e^{i\mu\pi}[1+\frac{1}{2i\mu}I(\pi)-\frac{1}{8\mu^2}I^2(\pi)]\\
&\quad -\frac{e^{-i\mu\pi}}{2i\mu}\int_0^\pi e^{2i\mu t}q(t)dt
 +o(1/\mu^2) e^{|Im\mu|\pi}\\
&\quad -[e^{-i\mu\pi}[1-\frac{1}{2i\mu}I(\pi)-\frac{1}{8\mu^2}I^2(\pi)]
-\frac{e^{i\mu\pi}}{2i\mu}\int_0^\pi e^{-2i\mu t}q(t)dt
 +o(1/\mu^2) e^{|Im\mu|\pi}]\Big\}.
\end{align*} %\label{e2.14}
Define $\Delta_0(\mu)=\frac{-(-1)^\theta
b}{2i\mu}(e^{i\pi\mu}-e^{-i\pi\mu})$. Combining like terms in the
above expression, gives
\begin{equation}
\begin{aligned}
\Delta(\mu)&=\Delta_0(\mu)-\frac{1}{2i\mu}[e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt
 -e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt]\\
&\quad +\frac{1}{4\mu^2}[e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt\int_0^t q(s)\,ds
 +e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt\int_0^t q(s)\,ds\\
&\quad -e^{i\pi\mu}\int_0^\pi q(t)dt\int_0^t e^{-2i\mu s}q(s)ds
 -e^{-i\pi\mu}\int_0^\pi q(t)dt\int_0^t e^{2i\mu s}q(s)ds]\\
&\quad -(-1)^\theta\frac{b}{2i\mu}\{\frac{e^{i\pi\mu}
 +e^{-i\pi\mu}}{2i\mu}I(\pi)
-\frac{e^{i\pi\mu}-e^{-i\pi\mu}}{8\mu^2}I^2(\pi)\\
&\quad -\frac{1}{2i\mu}(e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt
 +e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt)\}+o(1/\mu^3) e^{|Im\mu|\pi}.
\end{aligned}\label{e2.15}
\end{equation}


\section{Calculation of the regularized trace}

First, consider the case $\langle q\rangle=0$. Formula \eqref{e2.15} is then
noticeably simplified:
\begin{equation}
\Delta(\mu)=\Delta_0(\mu)(1+r(\mu)),\label{e3.1}
\end{equation}
where
\begin{equation}
\begin{aligned}
&r(\mu)\\
&=\frac{1}{\Delta_0(\mu)}
\Big\{-\frac{1}{2i\mu}[e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt
-e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt]\\
&\quad +\frac{1}{4\mu^2}\big[e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt\int_0^t q(s)\,ds
+e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt\int_0^t q(s)\,ds\\
&\quad -e^{i\pi\mu}\int_0^\pi q(t)dt\int_0^t e^{-2i\mu s}q(s)ds
-e^{-i\pi\mu}\int_0^\pi q(t)dt\int_0^t e^{2i\mu s}q(s)ds\big]
\\
&\quad +(-1)^{\theta+1}\frac{b}{4\mu^2}(e^{-i\pi\mu}\int_0^\pi
 e^{2i\mu t}q(t)dt+e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt)
+o(1/\mu^3) e^{|Im\mu|\pi}\Big\}.
\end{aligned}\label{e3.2}
\end{equation}
Integrating by parts the terms on the right-hand side of \eqref{e3.2},
we have
\begin{equation}
\begin{aligned}
r(\mu)&= \frac{(-1)^{\theta+1}}{4b\mu\sin\pi\mu}\{(e^{i\pi\mu}
+e^{-i\pi\mu})(q(\pi)-q(0))\\
&\quad -[e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q'(t)dt
+e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q'(t)dt]\}+\frac{q(\pi)+q(0)}{4\mu^2}
+o(1/\mu^2).
\end{aligned}\label{e3.3}
\end{equation}
Denote by $\Gamma_N$ the circle of radius $N+1/2$ centered at the
origin. It is well known \cite{m3} that the eigenvalues of
\eqref{e1.1}-\eqref{e1.2} form a sequence $\lambda_n=\mu_n^2$, where
$\mu_n=n+o(1)$, $n=1,2,\dots$. This asymptotic relation for the
eigenvalues implies that, for all sufficiently large $N$, the
numbers $\mu_n$ with $n\le N$ are inside $\Gamma_N$, and the
numbers $\mu_n$ with $n>N$
 are outside $\Gamma_N$. It follows that
$$
2\sum_{n=1}^N\mu_n^2=\frac{1}{2\pi
i}\oint_{\Gamma_N}\mu^2\frac{\Delta'(\mu)}{\Delta(\mu)}d\mu;
$$
see \cite{m3}.
Obviously, if $\mu\in\Gamma_N$, then $|\Delta_0(\mu)|\ge c_1e^{|Im\mu|\pi}/|\mu|$ $(c_1>0)$.
This inequality and the Riemann lemma \cite{m3} imply that
 $\max_{\mu\in\Gamma_N}|r(\mu)|\to0$ as
$N\to\infty$. Combining this with \eqref{e3.1} yields
\begin{equation}
\begin{aligned}
\frac{1}{2\pi i}\oint_{\Gamma_N}\mu^2\frac{\Delta'(\mu)}{\Delta(\mu)}d\mu
&= \frac{1}{2\pi i}\oint_{\Gamma_N}\mu^2
 \big(\frac{\Delta_0'(\mu)}{\Delta_0(\mu)}+\frac{r'(\mu)}{1+r((\mu)}\big)d\mu
\\
&= 2\sum_{n=1}^N n^2+\frac{1}{2\pi i}\oint_{\Gamma_N}\mu^2d\ln(1+r(\mu))\\
&= 2\sum_{n=1}^N n^2-\frac{1}{2\pi i}\oint_{\Gamma_N}2\mu\ln(1+r(\mu))d\mu.
\end{aligned}\label{e3.4}
\end{equation}
Expanding $\ln(1+r(\mu))$ by the Maclaurin formula and applying \eqref{e3.3}
and the Riemann lemma \cite{m3},
we find that
\begin{equation}
\ln(1+r(\mu))=
r(\mu)-\frac{(q(\pi)-q(0))^2}{2b^2\mu^2}\cot^2\pi\mu
+\frac{o(1)}{\mu^2}\label{e3.5}
\end{equation}
on $\Gamma_N$. Evidently,
\begin{equation}
\lim_{|Im\mu|\to\infty}(\cot^2\pi\mu+1)=0.\label{e3.6}
\end{equation}
It follows from \eqref{e3.5} and \eqref{e3.6} that
\begin{align*}
&\frac{1}{2\pi i}\oint_{\Gamma_N}2\mu\ln(1+r(\mu))d\mu\\
&=\frac{1}{2\pi i}\oint_{\Gamma_N}\big[
\frac{(-1)^{\theta+1}}{2b\sin\pi\mu}\{(e^{i\pi\mu}+e^{-i\pi\mu})(q(\pi)-q(0))
\\
&\quad -(e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q'(t)dt
+e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q'(t)dt)\}\\
&\quad -\frac{(q(\pi)-q(0))^2}{b^2\mu}cot^2\pi\mu+\frac{q(\pi)+q(0)}{2\mu}+\frac{o(1)}{\mu}\big]
d\mu\\
&=\sum_{n=-N}^N\Big[(-1)^{\theta+1}\big[(e^{i\pi n}+e^{-i\pi n})(q(\pi)-q(0))
-(e^{i\pi n}\int_0^\pi e^{-2int}q'(t)dt \\
&\quad +e^{-int}\int_0^\pi  \int e^{2int}q'(t)dt)\big]\Big]
\big/(2b\pi\cos\pi n) \\
&\quad +\frac{1}{2\pi
i}\oint_{\Gamma_N}[\frac{(q(\pi)-q(0))^2}{b^2\mu}
 +\frac{q(\pi)+q(0)}{2\mu}]d\mu+o(1)\\
&=\frac{2(-1)^{\theta+1}}{\pi b}\sum_{n=1}^N(q(\pi)-q(0)\\
&\quad -\int_0^\pi q'(t)\cos(2nt)dt)
+\frac{(q(\pi)-q(0))^2}{b^2}+\frac{q(\pi)+q(0)}{2}+o(1).
\end{align*}
Combining this and \eqref{e3.4}, we obtain
\begin{equation}
\begin{aligned}
2\sum_{n=1}^N\mu_n^2
&=2\sum_{n=1}^N n^2 +\frac{2(-1)^{\theta}}{\pi b}
  \sum_{n=1}^N(q(\pi)-q(0)-\int_0^\pi q'(t)\cos(2nt)dt)\\
&\quad -\frac{(q(\pi)-q(0))^2}{b^2}-\frac{q(\pi)+q(0)}{2}+o(1).
\end{aligned}\label{e3.7}
\end{equation}
Passing to the limit as $N\to\infty$ in \eqref{e3.7}, we have
\begin{equation}
\begin{aligned}
&\sum_{n=1}^\infty\big\{\lambda_n-n^2
-\frac{(-1)^{\theta}(q(\pi)-q(0)-\int_0^\pi q'(t)\cos(2nt)dt)}{\pi b}\big\}\\
&\quad +\frac{(q(\pi)-q(0))^2}{2b^2}+\frac{q(\pi)+q(0)}{4}=0.
\end{aligned}
\label{e3.8}
\end{equation}

Now consider the case $\langle q\rangle\ne0$. Let $\tilde q(x)=q(x)-\langle q\rangle$.
Then $\langle\tilde q\rangle=0$. Suppose that  \eqref{e1.1}-\eqref{e1.2}
with potential $\tilde q$ has eigenvalues $\tilde\lambda_n$. Then
$\tilde \lambda_n=\lambda_n-\langle q\rangle$. According to \eqref{e3.8}, we have
\begin{align*}
&\sum_{n=1}^\infty\big\{\tilde\lambda_n-n^2
-\frac{(-1)^{\theta}(\tilde q(\pi)-\tilde q(0)
-\int_0^\pi \tilde q'(t)\cos(2nt)dt)}{\pi b}\big\}\\
&\quad +\frac{(\tilde q(\pi)-\tilde q(0))^2}{2b^2}+\frac{\tilde{q}(\pi)+\tilde {q}(0)}{4}=0.
\end{align*}
Substituting the expressions for $\tilde\lambda_n$ and $\tilde q(x)$
into this equality, we obtain
formula \eqref{e1.3}.



\subsection*{Acknowledgments}
This work was supported by grant 07-01-00158 from the Russian Foundation
for Basic Research.


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\end{document}